[Confuse!]Capacitor and resistor in DC circuit, please help

The figure in attached file below shows a circuit in which a battery (electromotive force = 10V) is connected with three capacitors (capacitance = 1 microFarad, 2microFarad,3microFarad) and two resistors (resistance = 2Ω and 3Ω)

3. The attempt at a solution
I understand that capacitors are in this problem is fully charged so I can analyze them as open circuit. It means no current passes through three capacitors and I can solve this problem only voltage and resistance circuit

Thus, resistance 2Ω and 3Ω are connected in series and Rtotal = 5Ω
From ohm's law >> V=IR
I get Itotal = 10/5 = 2A
So, Voltage at 3Ω is V3Ω = 2*3 =6 V is the VQ

and VP is 10 V
Thus, I get potential difference is 4 V

However, the answer is 3V!!! What did I understand something wrong.. Please help I really stuck with this problem

Thanks a lot, for answer this is my first post on PF 1. The problem statement, all variables and given/known data

Staff: Mentor

The figure in attached file below shows a circuit in which a battery (electromotive force = 10V) is connected with three capacitors (capacitance = 1 microFarad, 2microFarad,3microFarad) and two resistors (resistance = 2Ω and 3Ω)

3. The attempt at a solution
I understand that capacitors are in this problem is fully charged so I can analyze them as open circuit. It means no current passes through three capacitors and I can solve this problem only voltage and resistance circuit

Thus, resistance 2Ω and 3Ω are connected in series and Rtotal = 5Ω
From ohm's law >> V=IR
I get Itotal = 10/5 = 2A
So, Voltage at 3Ω is V3Ω = 2*3 =6 V is the VQ

and VP is 10 V
Thus, I get potential difference is 4 V

However, the answer is 3V!!! What did I understand something wrong.. Please help I really stuck with this problem

Thanks a lot, for answer this is my first post on PF

Hi IIK*JII, Welcome to Physics Forums.

You've calculated the final potential at the node where the resistors meet (let's call it node R) to be 6 V, but that is not node Q where all the capacitors meet; they are distinct nodes.

Nodes Q and R are separated by a 2 μF capacitor which will pass no current at steady state, but you cannot assume that the potential on that capacitor at steady state will be zero.

Piont P isn't the same point as the point between the resistances, so it doesn't have to have the same potential as this point.

To solve this:

set the potential at point Q equal to V_Q.

Use V = qC for all three capacitors. You can compute all charges on the capacitors as a function of V_Q. You know all potentials for one of the sides of each capacitor.

If all the capacitors are initially uncharged, Kirchhoffs' current law: [itex] \Delta I = 0 [/itex] for the point Q, is valid for the charges on the capacitors as well, since I = dq/dt, so [itex] \Delta q = 0 [/itex] (where a charge on a capacitor is positive if the side of the capacitor connected to Q is more positive)

I have something wondering node R connected between 2 resistors (2Ω&3Ω) and because 3Ω
connect with capacitors so no current passes 3Ω ???,,, So, VR=V@2Ω??

Do not mix potential with potential difference. Point R is at a potential with respect to a reference point. There is a potential difference across the 2 ohm resistor which is equal the current through it multiplied by resistance.
The current flows where it can. The current flowing out from the positive terminal of the battery, can flow through the 2 ohm resistor, and reaching point R, flows further through the 3 ohm resistor toward the negative terminal of the battery, following the red line.

Yo need to operate with the charges on the capacitors. The three capacitor has got a common point, Q. The total charge on the connected plates must be zero: -q1+q2+q3=0.

The potential of point R is 6 V with respect to the negative terminal of the battery (which is the same as that of point O). If point Q is at potential U with respect O , the potential difference across the 1 μ F capacitor is U, that across the 3 μF capacitor is 10-U, and the PD across the 2μF capacitor is U-6. Find the charges in terms of U and use the charge neutrality at point Q.